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anonymous
 one year ago
I have the answer but i dont know how..
The lower edge of a mural, 12 ft high, is 6ft above an observer’s eye. Under the assumption that the
most favorable view is obtained when the angle subtended by the mural at the eye is maximum, at
what distance from the wall should the observer stand?
anonymous
 one year ago
I have the answer but i dont know how.. The lower edge of a mural, 12 ft high, is 6ft above an observer’s eye. Under the assumption that the most favorable view is obtained when the angle subtended by the mural at the eye is maximum, at what distance from the wall should the observer stand?

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433509816192:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we have the same drawing sir

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0From the lower small triangle \[\tan(s) = \frac{6}{x}\tag{1}\] From the big triangle \[\tan(t+s)=\frac{18}{x}\tag{2}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Our goal is to eliminate \(s\) somehow and get an equation with \(t\) and \(x\) as variables

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0In view of that, apply the angle sum formula : \[\large \tan(t)=\tan((t+s)s)=\frac{\tan(t+s)\tan(s)}{1+\tan(t+s)\tan(s)}\] plugin the values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok sir i will solve now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got this sir: dw:1433510419375:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0doesn't look correct

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\large{\begin{align} \tan(t)=\tan((t+s)s) &=\frac{\tan(t+s)\tan(s)}{1+\tan(t+s)\tan(s)}\\~\\ &=\dfrac{\dfrac{18}{x}\dfrac{6}{x}}{1+\dfrac{18}{x}\dfrac{6}{x}}\\~\\ &=\dfrac{12x}{x^2+108} \end{align}}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so the equation is \[\large \tan(t)=\dfrac{12x}{x^2+108}\] and we want to maximize \(t\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got that equation sir

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0may be isolate \(t\) and find \(\dfrac{dt}{dx}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is this the bases sir? dw:1433510949377:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \tan(t)=\dfrac{12x}{x^2+108}\] \[\large t = \tan^{1}\left(\dfrac{12x}{x^2+108}\right)\] \[\large \dfrac{dt}{dx} = ?\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0remember the derivative of \(\tan^{1}(x)\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433511150272:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\large t = \tan^{1}\left(\dfrac{12x}{x^2+108}\right)\] \[\large{\begin{align} \dfrac{dt}{dx} &= \dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\dfrac{d}{dx}\left(\dfrac{12x}{x^2+108}\right)\\~\\~\\ &=? \end{align}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok wait sir i will solve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 1+\frac{ 12X ^{2} }{ x ^{4}+216x ^{2}+11664 } }\times \frac{ 12x ^{2}+1296 }{ x ^{4}+216x ^{2}+11664 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh there an error im sorry.. it should be 144x^2 instead of having 12x^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i was stuck into this.. i dont know how to continue...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i already have an headache..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0because we will be setting all that equal to 0 and denominator gona vanish

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\large{\begin{align} \dfrac{dt}{dx} &= \dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\dfrac{d}{dx}\left(\dfrac{12x}{x^2+108}\right)\\~\\~\\ &=\dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\left(\dfrac{(x^2+108)1212x(2x)}{(x^2+108)^2}\right)\\~\\~\\ \end{align}}\] yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0setting that equal to 0 gives \[\large (x^2+108)1212x(2x) = 0\] solve \(x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but sir im a little bit confused by this... how come the remaining equation becomes \[(x ^{2}+108)1212x(2x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did the others vanish? and this is the equation that still remains?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where did this \[\frac{ 1 }{ 1 +\left( \frac{ 12x }{ x ^{2}+108 } \right)^{2}}\] go?
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